Question: Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2 - x}{x - 9} = \dfrac{4x + 36}{x - 9}$
Explanation: Multiply both sides by $x - 9$ $ \dfrac{x^2 - x}{x - 9} (x - 9) = \dfrac{4x + 36}{x - 9} (x - 9)$ $ x^2 - x = 4x + 36$ Subtract $4x + 36$ from both sides: $ x^2 - x - (4x + 36) = 4x + 36 - (4x + 36)$ $ x^2 - x - 4x - 36 = 0$ $ x^2 - 5x - 36 = 0$ Factor the expression: $ (x + 4)(x - 9) = 0$ Therefore $x = -4$ or $x = 9$ However, the original expression is undefined when $x = 9$. Therefore, the only solution is $x = -4$.